Chapter 3 Solution of Linear Systems - Math/CS

78 CHAPTER 3. SOLUTION OF LINEAR SYSTEMS
Problem 3.5.8. Implement the function gepp, and use it to solve the linear systems given in
problems 3.2.2 and 3.2.3.
Problem 3.5.9. Test gepp using the linear system
⎡
1 2
⎤ ⎡
3
⎣ 4 5 6 ⎦ ⎣
7 8 9
Explain your results.
x1
x2
x3
⎤
⎡
⎦ = ⎣
Problem 3.5.10. Modify gepp, replacing the statements if abs(A(k,k)) < tol and if abs(A(n,n))
< tol with, respectively, if A(k,k) == 0 and if A(n,n) == 0. Solve the linear system given in
problem 3.5.9. Why are these results different than those computed in problem 3.5.9?
Problem 3.5.11. The built-in Matlab function hilb can be used to construct the so-called Hilbert
matrix, whose (i, j) entry is 1/(i + j − 1). Write a script M-file that constructs a series of test
problems of the form:
A = hilb(n);
x_true = ones(n,1);
b = A*x_true;
The script should use gepp to solve the resulting linear systems, and print a table of results with the
following information:
where
----------------------------------------------------n
error residual condition number
-----------------------------------------------------
• error = relative error = norm(x true - x)/norm(x true)
• residual = relative residual error = norm(b - A*x)/norm(b)
• condition number = measure of conditioning = cond(A)
Print a table of results for n = 5, 6, . . . , 13. Are the computed residual errors small? What about the
relative errors? Is the size of the residual error related to the condition number? What about the
relative error? Now try to run the script with n ≥ 14. What do you observe?
Problem 3.5.12. Rewrite the function gepp so that it does not use array operations. Compare the
efficiency (e.g., using tic and toc) of this function with gepp on matrices of various dimensions.
3.5.4 Built-in Matlab Tools for Linear Systems
An advantage of using a powerful scientific computing environment like Matlab is that we do
not need to write our own implementations of standard algorithms, like Gaussian elimination and
triangular solves. Matlab provides two powerful tools for solving linear systems:
• The backslash operator: \
Given a matrix A and vector b, this operator can be used to solve Ax = b with the single
command:
x = A \ b;
1
0
1
⎤
⎦ .